Question

Magnetic field intensity at the centre of coil of $50$ turns, $0.5m$ radius and carrying a current of $2A$, is :

• A. $5\times 10^{-5}T$
• B. $3\times 10^{-5}T$
• C. $1.25\times 10^{-4}T$
• D. $0.5\times 10^{-5}T$

Answer 1

Answer: (C)

Magnetic field at the centre of a circular coil carrying current $i$ and of radius $a$, having $N$ number of turns is
$\therefore B=\frac{{\mu }_{0}Ni}{2a}$
Given, $N=50,$
$r=0.5m,$
$i=2A$
${\mu }_0=4\pi \times 10^{-7}Wb/Am$
$B=\frac{4\pi \times 10^{-7}}{2}\times \frac{50\times 2}{0.5}$
$=1.25\times 10^{-4}T$
Note : Direction of magnetic field is perpendicular to the plane of the coil, that is along the axis of the coil.